Going from Linear Equations to Linear Inequalities

This line is the picture of all the points $\,(x,y)\,$ that make the equation ‘$\,y=x+1\,$’ true.
How can ‘$\,y=x+1\,$’ be true? For a given $\,x$-value, the $\,y$-value must equal $\,x + 1\,$.
For each $\,x$-value, there is exactly one corresponding $\,y$-valuewhatever $\,x\,$ is, plus $\,1\,$.
The line is the picture of all the points $\,(x,x+1)\,$, as $\,x\,$ varies over all real numbers:
$$
\text{the line ‘$\,y=x+1\,$’ is all points of the form:} \qquad (\ \ x\ \ ,\overbrace{x+1}^{\text{the $y$-value EQUALS $x+1$}})
$$

graph of $\,y = x + 1\,$:
all points of the form $\,(x,x+1)$

Question:
What happens if the verb in the sentence ‘$\,y \color{red}{=} x+1\,$’ is changed from ‘$\,\color{red}{=}\,$’ to $\,\lt\,$, $\,\gt\,$, $\,\le\,$, or $\,\ge\,$?

Answer:
You go from a linear equation in two variables, to a linear inequality in two variables.
The solution set changes dramatically!
What was a line now becomes an entire half-plane:

graph of $\,y < x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is less than $\,x+1\,$

line is dashed;
shade below the line

graph of $\,y > x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is greater than $\,x+1\,$

line is dashed;
shade above the line

graph of $\,y \le x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is less than or equal to $\,x+1\,$

line is solid;
also shade below the line

graph of $\,y \ge x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is greater than or equal to $\,x+1\,$

line is solid;
also shade above the line

Important Concepts for Graphing Linear Inequalities in Two Variables

DEFINITION: LINEAR INEQUALITY IN TWO VARIABLES
A linear inequality in two variables is a sentence of
the form
$$ax + by + c < 0\,,$$
where $\,a\,$ and $\,b\,$ are not both zero; $\,c\,$ can be any real number.
The inequality symbol can be any of these: $\,\lt\,$, $\,\gt\,$, $\,\le\,$, or $\,\ge\,$
Remember: a ‘sentence of the form ...’ really means a ‘sentence that can be put in the form ...’

EXAMPLES OF LINEAR INEQUALITIES IN TWO VARIABLES

$3x - 4y + 5 > 0$

$y \le 5x - 1$

$x \ge 2\,$(shorthand for $\,x + 0y \ge 2\,$)

$y < 5\,$(shorthand for $\,0x + y < 5\,$

Key ideas for recognizing linear inequalities in two variables:

the verb must be an inequality symbol: $\,\lt\,$, $\,\le\,$, $\,\gt\,$, or $\,\ge\,$

the variables must be raised only to the first power :no squares, no variables in denominators, no variables under square roots, and so on

you don't need to have both $\,x\,$ and $\,y\,$, but you must have at least one of these variables

LINEAR INEQUALITIES GRAPH AS HALF-PLANES
Every linear inequality in two variables graphs
as a half-plane:

if the verb is $\,\lt\,$ or $\,\gt\,$, the boundary line is not included (dashed)

if the verb is $\,\le\,$ or $\,\ge\,$, the boundary line is included (solid)

WHICH HALF-PLANE TO SHADE?
If the linear inequality is in slope-intercept form (like $\,y < mx + b\,$), then it's easy to know which half of the line to shade:

if the sentence is $\,y \lt mx + b\,$ or $\,y \le mx + b\,$, shade BELOW the line

if the sentence is $\,y \gt mx + b\,$ or $\,y \ge mx + b\,$, shade ABOVE the line

This only works if the inequality is in slope-intercept form!
Of course, you can always put a sentence in slope-intercept form, by solving for $\,y\,$.
Then, you can use this method.
But, the ‘test point method’ (below) is usually quicker-and-easier, if the sentence isn't already in slope-intercept form.

The Test Point Method for Graphing Linear Inequalities in Two Variables

So, what about graphing something like $\,2x - y < 3\,$, which isn't
in slope-intercept form?
You can (if desired) solve for $\,y\,$, and then use the method above:
$$\begin{gather}
2x - y < 3\cr
-y < -2x + 3\cr
y > 2x - 3
\end{gather}
$$
(Remember to change the direction of the inequality symbol when you multiply/divide by a negative number.)

The graph of $\,2x - y < 3\,$ is the same as the graph of $\,y > 2x - 3\,$.
Graph the line $\,y = 2x - 3\,$ (dashed), and then shade everything above (see right).

However, there's an easier way. Keep reading!

graph of $\,2x - y < 3\,$
(which is equivalent to $\,y > 2x - 3\,$)

The ‘Test Point Method’ is so-called because it involves choosing a
‘test point’ to decide which side of the line to shade.
The process is illustrated with an example: graphing $\,2x - y < 3\,$.
The Test point Method is usually easiest to use with sentences that aren't in slope-intercept form.

GRAPH, USING THE TEST POINT METHOD: $\,2x - y < 3$

Step 1: IDENTIFICATION
Recognize that ‘$\,2x - y < 3\,$’ is a linear inequality in two variables.
Therefore, you know the graph is a half-plane.
You need the boundary line; you need to know which side to shade.

Step 2: BOUNDARY LINE
Graph the boundary line $\,2x - y = 3\,$ using the intercept method.
When $\,x = 0\,$, $\,y = -3\,$.
When $\,y = 0\,$, $\,x = \frac{3}{2}\,$.
Since the verb in ‘$\,2x - y < 3\,$’ is ‘$\,not included in the solution set.
Therefore, the line is dashed.

graph the boundary lineusing the intercept method

Step 3: TEST POINT TO DECIDE WHICH SIDE TO SHADE
Choose a simple point that is NOT on the line.
Whenever $\,(0,0)\,$ is available, choose it! Zeroes are very easy to work with.

Is $\,(0,0)\,$ in the solution set?
Substitute $\,x = 0\,$ and $\,y = 0\,$ into the original sentence, to see if it is true or false.
Put a question mark over the inequality symbol, since you're asking a question:
$$
2(0) - 0 \overset{?}{
If the result is FALSE, shade the other side.

Since ‘$\,0 < 3\,$’ is TRUE, shade the side containing $\,(0,0)\,$.
Done!

With so many zeroes involved in this method,computations
can often be done in your head, making this QUICK and EASY!

test point $\,(0,0)\,$;

$2(0) - 0 \overset{?}{
shade side containing test point

Special Linear Inequalities in Two Variables: You only see one variable

Out of context, sentences like ‘$\,x \ge 2\,$’ and ‘$\,y < 5\,$’ can be confusing.
You only see one variable, but that doesn't necessarily mean that there isn't another variable with a zero coefficient!

Out of context, here are clues to what is likely wanted:

‘Solve: $\,x\ge 2\,$’ probably wants you to treat this as an inequality in ONE variable.

‘Graph: $\,x\ge 2\,$’ probably wants you to treat this as an inequality in TWO variables.

As shown below, there is a BIG difference in the nature of the solution set!

graph of $\,x\ge 2\,$,
viewed as an inequality in ONE variable

Viewed as an inequality in one variable,
the solution set of ‘$\,x\ge 2\,$’ is
the set of all numbers that
are greater than or equal to $\,2\,$.

The solution set is the interval $\,[2,\infty)\,$, shown at left.

graph of $\,x\ge 2\,$,
viewed as an inequality in TWO variables

Viewed as an inequality in two variables,
‘$\,x\ge 2\,$’ is really a shorthand for ‘$\,x + 0y \ge 2\,$’.

The solution set is the set of all points $\,(x,y)\,$,
where the $x$-value is greater than or equal to $\,2\,$.
The $y$-value can be anything!

Here are examples of substitution into ‘$\,x + 0y \ge 2\,$’:
The point $\,(2,5)\,$ is in the solution set, since ‘$\,2 + 0(5) \ge 2\,$’ is TRUE.
The point $\,(3.5,-7.4)\,$ is in the solution set, since ‘$\,3.5 + 0(-7.4) \ge 2\,$’ is TRUE.

The graph is the half-plane shown at left.
This is the picture of all the points with $x$-value greater than or equal to $\,2\,$.

graph of $\,y\lt 5\,$,
viewed as an inequality in TWO variables

Viewed as an inequality in two variables,
‘$\,y\lt 5\,$’ is really a shorthand for ‘$\,0x + y \lt 5\,$’.

The solution set is the set of all points $\,(x,y)\,$,
where the $y$-value is less than $\,5\,$.
The $x$-value can be anything!

Here are examples of substitution into ‘$\,0x + y \lt 5\,$’:
The point $\,(2,4)\,$ is in the solution set, since ‘$\,0(2) + 4 \lt 5\,$’ is TRUE.
The point $\,(-7.4,-3)\,$ is in the solution set, since ‘$\,0(-7.4) -3 \lt 5\,$’ is TRUE.

The graph is the half-plane shown at left.
This is the picture of all the points with $y$-value less than $\,5\,$.

Master the ideas from this section
by practicing the exercise at the bottom of this page.